JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 15)
Interference fringes are observed on a screen
by illuminating two thin slits 1 mm apart with a
light source ($$\lambda $$ = 632.8 nm). The distance
between the screen and the slits is 100 cm. If
a bright fringe is observed on a screen at a
distance of 1.27 mm from the central bright
fringe, then the path difference between the
waves, which are reaching this point from the
slits is close is
1.27 $$\mu $$m
2.05 $$\mu $$m
2.87 nm
2 nm
Explanation
_2nd_September_Morning_Slot_en_15_1.png)
y = $${{nD\lambda } \over d}$$
$$ \Rightarrow $$ n = $${{yd} \over {D\lambda }}$$
= $${{1.27 \times {{10}^{ - 3}} \times {{10}^{ - 3}}} \over {1 \times 632.8 \times {{10}^{ - 9}}}}$$ = 2
Path difference $$\Delta $$x = n$$\lambda $$
= 2 $$ \times $$ 632.8 nm
= 1265.6 nm
= 1.27 $$\mu $$m
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